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Old 2009-09-16, 21:04   #7
maxal
 
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Feb 2005

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Quote:
Originally Posted by grandpascorpion View Post
Background: I'm trying to create an a.p. of 6 or more terms. (http://www.primepuzzles.net/puzzles/puzz_413.htm)
That's a tough problem.
If n is such that for some k of its divisors: d_1, d_2, \dots, d_k, we have
\frac{n}{d_i} + d_i = m + q\cdot i for i=1,2,\dots,k
then
(m+qi)^2 - 4n = \left( \frac{n}{d_i} - d_i \right)^2 for i=1,2,\dots,k
form a sequence of k squares whose second differences equal the constant 2 q^2.

For example, n=36400 gives a sequence of squares
33^2, 150^2, 213^2, 264^2, 309^2
whose second differences equal 2\cdot 27^2 = 1458.

Finding sequences of squares with constant second differences is a rather hard task (see the attached paper) and additional requirement of having difference of the special form 2 q^2 makes it even harder.
Attached Files
File Type: pdf browkin8572.pdf (146.3 KB, 212 views)

Last fiddled with by maxal on 2009-09-16 at 21:24
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